template base class typedef members invisible

问题

I'm aware of the fact that the 'dependent names' are not visible to the compiler by default. But I was told in answers to other SO questions (here, here, and ultimately on the C++ faq) that a using declaration may help.

So I tried.

A template base class:

// regardless of the fact that members are exposed...
template<typename T>
struct TBase {
   typedef T MemberType;
   MemberType baseMember;
   MemberType baseFunction() { return MemberType(); }
};

And a derived class, using the base's members:

template<typename T>
struct TDerived : public TBase<T> {
   // http://www.parashift.com/c++-faq-lite/nondependent-name-lookup-members.html
   // tells us to use a `using` declaration.
   using typename TBase<T>::MemberType;
   using TBase<T>::baseFunction;
   using TBase<T>::baseMember;

   void useBaseFunction() { 
      // this goes allright.
      baseFunction();
      ++baseMember;

      // but here, the compiler doesn't want to help...
      MemberType t; //error: expected `;' before ‘t’
   }
};

I tried this out on ideone. It has gcc-4.3.3 and gcc-4.5.1

Is this expected behavior? How are we supposed to work around the 'dependent name' law for accessing parent template class' member typedefs?

回答1:

You probably want to do:

using MemberType = typename TBase<T>::MemberType; // new type alias syntax

typedef typename TBase<T>::MemberType MemberType; // old type alias syntax

The syntax using Base::member; can only be used to bring the declarations of non-type members into scope.

Also note that none of these are actually required, you can qualify each use (for types with the base, for non-types with either this-> or the base) and that will make the symbol dependent.

来源：https://stackoverflow.com/questions/13405715/template-base-class-typedef-members-invisible

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